I am doing some research on ellipsoids. I am not sure where the formula for the surface area of a prolate ellipsoid comes from. Can anyone please help me with how to derive the formula. I have the formula below
[Math] Deriving formula for surface area of an ellipsoid
areaellipsoidsgeometrymultivariable-calculus
Related Solutions
In this answer, it is shown geometrically that the area of a strip of sphere between two parallels of latitude is the same as the area of the orthogonal projection of that strip onto the cylinder with the same radius as the sphere and whose axis is parallel to the north-south axis of the sphere.
Finding the area of this cylinder is just multiplying its height, $2r$, by its circumference, $2\pi r$. Thus, the area of the sphere is $4\pi r^2$.
Your son's procedure seems to be summing the circumferences of the green strips on the sphere, but there must be some accounting for the widths of these strips. This seems to be the problem he is encountering.
If $c$ is a regular value of a smooth function $f$, then $S=f^{-1}(c)$ is a surface for which $\nabla f(p)$ is normal to $T_pS$. Here $f(x,y,z) = 2x^2+y^2+2z^2$ and $p = (1,1,1)$, so the normal to the ellipsoid at $p$ is $$\nabla f(1,1,1) = (4x,2y,4z)\big|_{x=y=z=1} = (4,2,4).$$
Best Answer
The method is very standard and appears in most calculus texts.
Let $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be the ellipse such that $a>b$.
\begin{align*} y &= \frac{b}{a} \sqrt{a^2-x^2} \\ \frac{dy}{dx} &= -\frac{bx}{a\sqrt{a^2-x^2}} \\ ds &= \sqrt{1+\left( \frac{dy}{dx} \right)^2} \, dx \\ &= \sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}} \, dx \\ &= a\frac{\sqrt{1-\left( 1-\frac{b^2}{a^2} \right) \frac{x^2}{a^2}}} {\sqrt{a^2-x^2}} \, dx \\ S &= \int_{-a}^{a} 2\pi y \, ds \\ &= 4b\pi \int_{0}^{a} \sqrt{1-\left( 1-\frac{b^2}{a^2} \right)\frac{x^2}{a^2}} \, dx \\ &= 4b\pi \left[ \frac{x}{2} \sqrt{1-\left( 1-\frac{b^2}{a^2} \right) \frac{x^2}{a^2}}+ \frac{a^2}{2\sqrt{a^2-b^2}} \sin^{-1} \frac{x\sqrt{a^2-b^2}}{a^2} \right]_{0}^{a} \\ &= 2\pi b \left( b+\frac{a^2}{\sqrt{a^2-b^2}} \sin^{-1} \frac{\sqrt{a^2-b^2}}{a} \right) \\ \end{align*}
\begin{align*} x &= \frac{a}{b} \sqrt{b^2-y^2} \\ \frac{dx}{dy} &= -\frac{ay}{b\sqrt{b^2-y^2}} \\ ds &= \sqrt{1+\left( \frac{dx}{dy} \right)^2} \, dy \\ &= \sqrt{1+\frac{a^2y^2}{b^2(b^2-y^2)}} \, dy \\ &= b\frac{\sqrt{1+\left( \frac{a^2}{b^2}-1 \right) \frac{y^2}{b^2}}} {\sqrt{b^2-y^2}} \, dy \\ S &= \int_{-b}^{b} 2\pi x \, ds \\ &= 4a\pi \int_{0}^{b} \sqrt{1+\left( \frac{a^2}{b^2}-1 \right)\frac{y^2}{b^2}} \, dy \\ &= 4a\pi \left[ \frac{y}{2} \sqrt{1+\left( \frac{a^2}{b^2}-1 \right) \frac{y^2}{b^2}}+ \frac{b^2}{2\sqrt{a^2-b^2}} \sinh^{-1} \frac{y\sqrt{a^2-b^2}}{b^2} \right]_{0}^{b} \\ &= 2\pi a \left( a+\frac{b^2}{\sqrt{a^2-b^2}} \sinh^{-1} \frac{\sqrt{a^2-b^2}}{b} \right) \\ \end{align*}